The RD Sharma Class 9 solutions for the chapter “Circle” exercise 16.5 is given here. In this exercise, students will learn about cyclic quadrilateral. A quadrilateral is a cyclic quadrilateral if its all vertices lie on a circle. This exercise includes several questions to help the students learn more about circles. Students are advised to click on the link below and get pdf to practice more on topic – circle. RD Sharma Solutions class 9 chapter 16 to help students in their endeavor to have a better understanding of circles.
Access Answers to Maths RD Sharma Class 9 Chapter 16 Circles Exercise 16.5 Page number 16.83
Question 1: In figure, ΔABC is an equilateral triangle. Find m∠BEC.
ΔABC is an equilateral triangle. (Given)
Each angle of an equilateral triangle is 60 degrees.
In quadrilateral ABEC:
∠BAC + ∠BEC = 180o (Opposite angles of quadrilateral)
60o + ∠BEC = 180 o
∠BEC = 180 o – 60 o
∠BEC = 120 o
Question 2: In figure, Δ PQR is an isosceles triangle with PQ = PR and m∠PQR=35°. Find m∠QSR and m∠QTR.
Given: ΔPQR is an isosceles triangle with PQ = PR and m∠PQR = 35°
∠PQR = ∠PRQ = 35o (Angle opposite to equal sides)
Again, by angle sum property
∠P + ∠Q + ∠R = 180 o
∠P + 35 o + 35 o = 180 o
∠P + 70 o = 180 o
∠P = 180 o – 70 o
∠P = 110 o
Now, in quadrilateral SQTR,
∠QSR + ∠QTR = 180 o (Opposite angles of quadrilateral)
110 o + ∠QTR = 180 o
∠QTR = 70 o
Question 3: In figure, O is the centre of the circle. If ∠BOD = 160o, find the values of x and y.
From figure: ∠BOD = 160 o
By degree measure theorem: ∠BOD = 2 ∠BCD
160 o = 2x
or x = 80 o
Now, in quadrilateral ABCD,
∠BAD + ∠BCD = 180 o (Opposite angles of Cyclic quadrilateral)
y + x = 180 o
Putting value of x,
y + 80 o = 180 o
y = 100 o
Answer: x = 80 o and y = 100 o.
Question 4: In figure, ABCD is a cyclic quadrilateral. If ∠BCD = 100o and ∠ABD = 70o, find ∠ADB.
In quadrilateral ABCD,
∠DCB + ∠BAD = 180o (Opposite angles of Cyclic quadrilateral)
100 o + ∠BAD = 180o
∠BAD = 800
In Δ BAD:
By angle sum property: ∠ADB + ∠DAB + ∠ABD = 180 o
∠ADB + 80o + 70 o = 180 o
∠ADB = 30o
Question 5: If ABCD is a cyclic quadrilateral in which AD||BC (figure). Prove that ∠B = ∠C.
Given: ABCD is a cyclic quadrilateral with AD ‖ BC
=> ∠A + ∠C = 180o ………(1)[Opposite angles of cyclic quadrilateral]
and ∠A + ∠B = 180o ………(2)[Co-interior angles]
Form (1) and (2), we have
∠B = ∠C
Question 6: In figure, O is the centre of the circle. Find ∠CBD.
Given: ∠BOC = 100o
By degree measure theorem: ∠AOC = 2 ∠APC
100 o = 2 ∠APC
or ∠APC = 50 o
∠APC + ∠ABC = 180 o (Opposite angles of a cyclic quadrilateral)
50o + ∠ABC = 180 o
or ∠ABC = 130 o
Now, ∠ABC + ∠CBD = 180 o (Linear pair)
130o + ∠CBD = 180 o
or ∠CBD = 50 o
Question 7: In figure, AB and CD are diameters of a circle with centre O. If ∠OBD = 500, find ∠AOC.
Given: ∠OBD = 500
Here, AB and CD are the diameters of the circles with centre O.
∠DBC = 900 ….(i)[Angle in the semi-circle]
Also, ∠DBC = 500 + ∠OBC
900 = 500 + ∠OBC
or ∠OBC = 400
Again, By degree measure theorem: ∠AOC = 2 ∠ABC
∠AOC = 2∠OBC = 2 x 400 = 800
Question 8: On a semi-circle with AB as diameter, a point C is taken, so that m(∠CAB) = 300. Find m(∠ACB) and m(∠ABC).
Given: m(∠CAB)= 300
To Find: m(∠ACB) and m(∠ABC).
∠ACB = 900 (Angle in semi-circle)
In △ABC, by angle sum property: ∠CAB + ∠ACB + ∠ABC = 1800
300 + 900 + ∠ABC = 1800
∠ABC = 600
Answer: ∠ACB = 900 and ∠ABC = 600
Question 9: In a cyclic quadrilateral ABCD if AB||CD and B = 70o , find the remaining angles.
A cyclic quadrilateral ABCD with AB||CD and B = 70o.
∠B + ∠C = 180o (Co-interior angle)
700 + ∠C = 1800
∠C = 1100
=> ∠B + ∠D = 1800 (Opposite angles of Cyclic quadrilateral)
700 + ∠D = 1800
∠D = 1100
Again, ∠A + ∠C = 1800 (Opposite angles of cyclic quadrilateral)
∠A + 1100 = 1800
∠A = 700
Answer: ∠A = 700 , ∠C = 1100 and ∠D = 1100
Question 10: In a cyclic quadrilateral ABCD, if m ∠A = 3(m∠C). Find m ∠A.
∠A + ∠C = 180o …..(1)[Opposite angles of cyclic quadrilateral]
Since m ∠A = 3(m∠C) (given)
=> ∠A = 3∠C …(2)
Equation (1) => 3∠C + ∠C = 180 o
or 4∠C = 180o
or ∠C = 45o
From equation (2)
∠A = 3 x 45o = 135o
Question 11: In figure, O is the centre of the circle ∠DAB = 50°. Calculate the values of x and y.
Given : ∠DAB = 50o
By degree measure theorem: ∠BOD = 2 ∠BAD
so, x = 2( 500) = 1000
Since, ABCD is a cyclic quadrilateral, we have
∠A + ∠C = 1800
500 + y = 1800
y = 1300
RD Sharma Solutions for Class 9 Maths Chapter 16 Exercise 16.5
RD Sharma Solutions Class 9 Maths Chapter 16 Circles Exercise 16.5 is based on the following topics and subtopics:
- Cyclic Quadrilateral
- Important results:
-The opposite angles of a Cyclic Quadrilateral are supplementary.
– The quadrilateral formed by angle bisectors of a cyclic quadrilateral is also cyclic.
– A cyclic trapezium is isosceles and its diagonals are equal.
– If two opposite sides of a cyclic quadrilateral are equal, then the other two sides are parallel.